Crystallographic Point Groups -

INTERNATIONAL SCHOOL ON FUNDAMENTAL CRYSTALLOGRAPHY

CRYSTALLOGRAPHIC POINT GROUPS

Bilbao Crystallographic Server

http://www.cryst.ehu.es

Cesar Capillas, UPV/EHU 1

Mois I. AroyoUniversidad del Pais Vasco, Bilbao, Spain

(short review)

The equilateral triangle allows six symmetry operations: rotations by 120 and 240 around its centre, reflections through the three thick lines intersecting the centre, and the identity operation.

1. Crystallographic symmetry operations

Symmetry operations of an object

The isometries which map the object onto itself are called symmetry operations of this object. The symmetry of the object is the set of all its symmetry operations.

If the object is a crystal pattern, representing a real crystal, its symmetry operations are called crystallographic symmetry operations.

Crystallographic symmetry operations

The symmetry operations are isometries, i.e. they are special kind of mappings between an object and its image that leave all distances and angles invariant.

GROUP THEORY(few basic facts)

Crystallographic symmetry operations in the plane

my

Mirror line my at 0,y

x

y=

-x

y =-1

1

x

y

det -1

1= ? tr -1

1= ?

Matrix representation

Fixed points

myxf

yf=

xf

yf

Mirror symmetry operation

drawing: M.M. JulianFoundations of Crystallography

Taylor & Francis, 2008c�

2-fold rotation

2zx

y=

-x

-x =-1

-1

x

y

det -1

-1=

tr -1

-1= ?

Symmetry operations in the planeMatrix representations

3+ x

y=

-y

x-y =0 -1

1 -1

x

y

det =

tr = ?

Matrix representation

0 -1

1 -1

0 -1

1 -1

??

3-fold rotation

GROUP AXIOMS

1. CLOSURE

2. IDENTITY

3. INVERSE ELEMENT

4. ASSOCIATIVITY

g1o g2=g12 g1, g2, g12∈G

g o e = e o g = g

g o g-1= e

(g1o g2)o g3= g1o ( g2 o g3)= g1o g2 o g3

Group properties and presentation

1. Order of a group

2. Multiplication table

3. Group generators

a set of elements such that each element of the group can be obtained as a product of the generators

number of elements

Group Properties

Multiplication table

Group generators a set of elements such that each element of the group can be obtained as a product of the generators

4. How to define a group

G G’

.. ..

G={g} G’={g’}Φ(g)=g’

Φ-1: G’ G

Φ(g1)Φ(g2)= Φ(g1g2)homomorphic condition

Isomorphic groups

Φ-1(g’)=g

Φ: G G’

-groups with the same multiplication table

g1g2

.g1g2

Φ(g1)

Φ(g2)

Φ(g1g2).

Crystallographic Point Groups in 2D

Point group 2 = {1,2}-group axioms?

-order of 2?

-multiplication table

-1

-1

-1

-1

1

1=x2 x 2 =

Where is the two-fold point?

Motif with symmetry of 2

-generators of 2?




Point group m = {1,m}

-group axioms?

-order of m?


-1

1

-1

1

1

1=xm x m =

Whereis the mirror line?

Motif with symmetry of m

-generators of m?drawing: M.M. Julian

Foundations of CrystallographyTaylor & Francis, 2008c�

GG’.. ..

Φ(g1)Φ(g2)= Φ(g1 g2)

Isomorphic groups

-groups with the same multiplication table

Point group 2 = {1,2}

Φ(g)=g’

Φ-1(g’)=g

Point group m = {1,m}


Point group 1 = {1}

-group axioms?

-order of 1?


1

1

1

1

1

1=x1 x 1 =

Motif with symmetry of 1

-generators of 1?



Problem 2.1

Consider the model of the molecule of the organic semiconductor pentacene (C22H14):

-symmetry operations: matrix and (x,y) presentation

-multiplication table-generators

Determine:

Problem 2.2

Consider the symmetry group of the square. Determine:



-generators

Visualization of Crystallographic Point Groups

- general position diagram - symmetry elements diagram

Stereographic Projections

Points P in the projection plane

P’’

EXAMPLE

general position symmetry elements

Stereographic Projections of mm2

Molecule of pentacene

Point group mm2 = {1,2z,mx,my}

Stereographic projections diagrams

Problem 2.2 (cont.)

general position symmetry elements

Stereographic Projections of 4mm

diagram diagram

? ?

Problem 2.3 (additional)

Consider the symmetry group of the equilateral triangle. Determine:



-generators

-general-position and symmetry-elements stereographic projection diagrams;

Conjugate elements

Conjugate elements gi ~ gk if ∃ g: g-1gig = gk,where g, gi, gk, ∈ G

Classes of conjugate elements

L(gi)={gj| g-1gig = gj, g∈G}

Conjugation-properties

(iii)

(i) L(gi) ∩ L(gj) = {∅}, if gi ∉ L(gj)

(ii) |L(gi)| is a divisor of |G|

(iv) if gi, gj ∈ L, then (gi)k=(gj)k= e

L(e)={e}

Example (Problem 2.2):

The group of the square 4mm

Classes of conjugate elements: {1}, {2},{4,4-},{mx,my}, {m+,m-}

Classes of conjugate elements

Problem 2.1 (cont)EXERCISES

Distribute the symmetry elements of the group mm2 = {1,2z,mx,my} in classes of conjugate elements.

multiplication table

stereographic projection

CRYSTALLOGRAPHIC POINT GROUPS(brief overview)

Symmetry operations in 3DRotations

Rotation Crystallographic Point Groups in 3D

Cyclic: 1(C1), 2(C2), 3(C3), 4(C4), 6(C6)

Dihedral: 222(D2), 32(D3), 422(D4), 622(D6)

Cubic: 23 (T), 432 (O)

Dihedral Point Groups

{e,2z, 2y,2x}

{e,3z,3z ,21,22,23}

{e,4z,4z, 2z,2y,2x,2+,2-}

{e,6z,6z, 3z,3z, 2z

21,22,23, 21,22,23}

222

32

422

622

´ ´ ´

Cubic Rotational Point Groups

23 (T)

{e, 2x, 2y, 2z, 31,31,32,32,33,33,34,34}

{e, 2x, 2y, 2z,4x,4x,4y,4y,4z,4z

31,31,32,32,33,33,34,34

21,22,23,24,25,26}

432(O)

Symmetry operations in 3DRotoinvertions

Symmetry operations in 3DRotoinvertions

Symmetry operations in 3DRotoinversions

Symmetry operations in 3D3 Roto-inversion

generalview

down the symmetry axis

3+generalview

3+down thesymmetry axis


Proper rotations: det =+1: 1, 2, 3, 4, 6

Improper rotations: det =-1:-2

-3

-4

-6

-1

Hermann-Mauguin symbolism (International Tables A)

-symmetry elements in decreasing order of symmetry (except for two cubic groups: 23 and m ) -

3

-symmetry elements along primary, secondary and ternary symmetry directions

rotations: by the axes of rotationplanes: by the normals to the planes

Crystal systems and Crystallographic point groups

primary secondary ternary

Crystal systems and Crystallographic point groups

primary secondary ternary


The group G= is called a direct-product group

Direct-product groups

G1 x G2 {(g1,e2), g1∈G1} ≅ G1

Let G1 and G2 are two groups. The set of all pairs {(g1,g2), g1∈G1, g2∈G2} forms a group with respect to the product: (g1,g2) (g’1,g’2)= (g1g’1, g2g’2).

G1 x G2

Properties of

G1 x G2

G1 x G2

G1 x G2 {(e1,g2), g2∈G2} ≅ G2

{(e1,g2), g2∈G1}= {(e1,e2)}{(g1,e2), g1∈G1} ∩

∀ (g1,g2)∈ G1 x G2 : (g1,g2)= (g1,e2) (e1,g2)

(i)

(ii)

(iii)

G1 x {1,1}=G1+1.G1

Examples: Direct product groups

G2={1,1} group of inversion

Point group mm2 = {1,2z,mx,my}G1={1,2z} G2={1,mx}G1 x G2= {1.1, 2z.1, 1.mx, 2zmx=my}

Centro-symmetrical groupsG1: rotational groups

Crystallographic point groups and abstract groups

Cyclic groupsAbelian non-cyclic groups

Non-Abelian groups

Direct products of non-Abelian groups with cyclic groups of order 2

Crystal classes

Affine equivalence Abstract

groups

32 18

Abstractgroups

Crystallographic point groups

as abstract groups

Crystallographic Point Groups

G G+1G G(G’) G’+1(G-G’)

1 (C1) 1+1.1=1 (Ci) ---- -----

2 (C2) 2+1.2=2/m (C2h) 2(1) m (Cs)

3 (C3) 3+1.3=3 (C3i or S6) ---- ----

4 (C4) 4+1.4=4/m (C4h) 4(2) 4 (S4)

6 (C6) 6+1.6=6/m (C6h) 6(3) 6 (C3h)


G G+1G G(G’) G’+1(G-G’)

222 (D2) 222+1.222=2/m2/m2/m 222(2) 2mm (C2v)

4/mmm(D4h) 422(222) 42m (D2d)

32 (D3) 32+1.32=32/m 3m(D3d) 32(3) 3m (C3v)

422 (D4) 422+1.422=4/m2/m2/m 422(4) 4mm (C4v)

6/mmm(D6h) 622(32) 62m (D3h) 622 (D6) 622+1.622=6/m2/m2/m 622(6) 6mm (C6v)

23 (T) 23+1.23=2/m3 m3 (Th) ---- -----

mmm (D2h)

432 (O) 432+1.432=4/m32/m 432(23) 43m (Td) m3m(Oh)


422 e 4z 4z 2z 2x 2y 2+2-

4mm e 4z 4z 2z mx my m+m-

42m e 4z 4z 2z 2x 2y m+m-

4m2 e 4z 4z 2z mx my 2+2-

Groups isomorphic to 422

Groups isomorphic to 622

622 e 6z6z 3z3z 2z 212223 212223´´ ´

6mm e 6z6z 3z3z 2z m1m2m3 m1m2m3´ ´ ´62m e 6z6z 3z3z mz 212223 m1m2m3´ ´ ´6m2 e 6z6z 3z3z mz m1m2m3 212223´ ´ ´

Problem 2.4

Consider the following three pairs of stereographic projections. Each of them correspond to a crystallographic point group isomorphic to 4mm:

(i) Determine those point groups by indicating their symbols, symmetry operations and possible sets of generators;(ii) Construct the corresponding multiplication tables;(iii) For each of the isomorphic point groups indicate the one-to-one correspondence with the symmetry operations of 4mm.

4mm


?xy

z

AB4 type molecule

Determine the symmetry group of the configuration:

Example Symmetry groups of molecules

Hint: cubic crystal systemsymmetry directions: [100] [111][110]

Problem 2.5(a)

Determine the symmetry elements and the corresponding point groups for each of the following models of molecules:

Problem 2.5(b)

Determine the symmetry elements and the corresponding point groups for each of the following models of molecules:

I. Subgroups: index, coset decomposition and normal subgroups

II. Conjugate subgroups

III. Group-subgroup graphs

GROUP-SUBGROUP RELATIONS

Subgroups: Some basic results (summary)

Subgroup H < G

1. H={e,h1,h2,...,hk} ⊂ G2. H satisfies the group axioms of G

Proper subgroups H < G, and trivial subgroup: {e}, G

Index of the subgroup H in G: [i]=|G|/|H| (order of G)/(order of H)

Maximal subgroup H of GNO subgroup Z exists such that:

H < Z < G

EXERCISES

Problem 2.6 Consider the group of the square and determine its subgroups

Multiplication table of 3m

mx

my

mxxx

y

Problem 2.7 (additional)

(ii) Distribute the subgroups into classes of conjugate subgroups;

(i) Consider the group of the equilateral triangle and determine its subgroups;

(iii) Construct the maximal subgroup graph of 3m

Coset decomposition G:H

Group-subgroup pair H < G

left coset decomposition

right coset decomposition

G=H+g2H+...+gmH, gi∉H, m=index of H in G

G=H+Hg2+...+Hgm, gi∉Hm=index of H in G

Coset decomposition-properties

(i) giH ∩ gjH = {∅}, if gi ∉ gjH

(ii) |giH| = |H|

(iii) giH = gjH, gi ∈ gjH

Theorem of Lagrange

group G of order |G|subgroup H<G of order |H|

then|H| is a divisor of |G|and [i]=|G:H|

Corollary The order k of any element of G,gk=e, is a divisor of |G|

Normal subgroups

Hgj= gjH, for all gj=1, ..., [i]

Coset decomposition G:H

EXERCISES

Problem 2.9 Demonstrate that H is always a normal subgroup if |G:H|=2.

Problem 2.8

Consider the subgroup {e,2} of 4mm, of index 4:

-Write down and compare the right and left coset decompositions of 4mm with respect to {e,2};

-Are the right and left coset decompositions of 4mm with respect to {e,2} equal or different? Can you comment why?

Conjugate subgroups

Conjugate subgroups Let H1<G, H2<G

then, H1 ~ H2, if ∃ g∈G: g-1H1g = H2

(i) Classes of conjugate subgroups: L(H)

(ii) If H1 ~ H2, then H1 ≅ H2

(iii) |L(H)| is a divisor of |G|/|H|

Normal subgroup

H G, if g-1H g = H, for ∀g∈G

EXERCISES

Problem 2.6 (cont) Distribute the subgroups of the group of the square into classes of conjugate subgroups

Hint: The stereographic projections could be rather helpful

Complete and contracted group-subgroup graphs

Complete graph of maximal subgroups

Contracted graph of maximal subgroups

Group-subgroup relations of point groupsInternational Tables for Crystallography, Vol. A, Chapter 10

Hahn and Klapper

Factor group

product of sets: Kj={gj1,gj2,...,gjn}Kk={gk1,gk2,...,gkm}

Kj Kk={ gjpgkq=gr | gjp ∈ Kj, gkq ∈Kk} Each element gr is taken only once in the product Kj Kk

G={e, g2, ...,gp} {

factor group G/H: H GG=H+g2H+...+gmH, gi∉H, G/H={H, g2H, ..., gmH}

(i) (giH)(gjH) = gijH(ii) (giH)H =H(giH)= giH

(iii) (giH)-1 = (gi-1)H

group axioms:

Problem 2.8 (cont)

Consider the normal subgroup {e,2} of 4mm, of index 4, and the coset decomposition 4mm: {e,2}:

(3) Show that the cosets of the decomposition 4mm:{e,2} fulfil the group axioms and form a factor group

(4) Multiplication table of the factor group

(5) A crystallographic point group isomorphic to the factor group?

GENERATION OF CRYSTALLOGRAPHIC POINT

GROUPS

Generation of point groups

Set of generators of a group is a set of group elements such that each element of the group can be obtained as an ordered product of the generators

g1 - identityg2, g3, ... - generate the rest of elements

Composition series: 1 Z2 Z3 ... Gindex 2 or 3

Crystallographic groups are solvable groups

W=(gh) * (gh-1) * ... * (g2) * g1kh kh-1 k2

Example Generation of the group of the square

Composition series: 1 2 4 4mmStep 1:

1 ={1}

Step 2: 2 = {1} + 2z {1}

Step 3: 4 ={1,2} + 4z {1,2}

Step 4: 4mm = 4 + mx 4

[2] [2] [2]

2z 4z mx

Generation of sub-cubic point groups

Composition series of cubic point groups and their subgroups

Generation of sub-hexagonal point groups

Composition series of hexagonal point groups and their subgroups

Problem 2.10

Generate the symmetry operations of the group 4/mmm following its composition series.

Generate the symmetry operations of the group 3m following its composition series.

GENERAL AND SPECIAL WYCKOFF POSITIONS

WXo = Xo

Site-symmetry group So={W} of a point Xo

General and special Wyckoff positions

=

General position Xo S= 1 ={1}

Special position Xo S> 1 ={1,...,}

a b c

d e f

g h i

x0

y0

z0

x0

y0

z0

Site-symmetry groups: oriented symbols


Point group 2 = {1,2z}

WXo = Xo

Site-symmetry group So={W} of a point Xo=(0,0,z)

=2z: 0

0

z

-1

-1

1

0

0

z

So = 2

2 b 1 (x,y,z) (-x,-y,z)

1 a 2 (0,0,z)

Example


Point group mm2 = {1,2z,mx,my}

WXo = Xo

Site-symmetry group So={W} of a point Xo=(0,0,0)

=2z:

=my:

0

0

z

-1

-1

1

1

-1

1

0

0

z

0

0

z

0

0

z

So = mm2

2 b .m. (x,0,z) (-x,0,z)

2 c m.. (0,y,z) (0,-y,z)

4 d 1 (x,y,z) (-x,-y,z) (x,-y,z) (-x,y,z)

1 a mm2 (0,0,z)

Example

Problem 2.11EXERCISES

Consider the symmetry group of the square 4mm and the point group 422 that is isomorphic to it.

Determine the general and special Wyckoff positions of the two groups.


EXAMPLE

Group-subgroup pair mm2 >2, [i]=2

mm2

x1,y1,z1 2 b 1

Wyckoff positions splitting schemes

4 d 1 (x,y,z) (-x,-y,z) (x,-y,z) (-x,y,z)

-x1,-y1,z1

x,-y,z=x2,y2,z2 2 b 1 -x,y,z=-x2,-y2,z2

x,y,z=-x,-y,z=

2

Problem 2.12EXERCISES

Consider the general and special Wyckoff positions of the symmetry group of the square 4mm and those of its subgroup mm2 of index 2.

Determine the splitting schemes of the general and special Wyckoff positions for 4mm > mm2.


GROUP-SUPERGROUPRELATIONS

Supergroups: Some basic results (summary)

Supergroup G>H

H={e,h1,h2,...,hk} ⊂ G

Proper supergroups G>H, and trivial supergroup: H

Index of the group H in supergroup G: [i]=|G|/|H| (order of G)/(order of H)

Minimal supergroups G of H

NO subgroup Z exists such that: H < Z < G

The Supergroup Problem

Given a group-subgroup pair G>H of index [i]

Determine: all Gk>H of index [i], Gi≃G

H

G G2 G3 Gn...

all Gk>H contain H as subgroup

H

G

[i] [i]

Gk=H+g2H+...+gikH

Example: Supergroup problem

Group-subgroup pair422>222

222

422

[2]

Supergroups 422 of the group 222

222

How many are the subgroups 222 of 422?

422 ?How many are

the supergroups 422 of 222?

Example: Supergroup problem

Group-subgroup pair422>222

2z2x2y

422

[2]

Supergroups 422 of the group 222

222

4z22

2z2+2-

4x22 4y22

[2]

4z22=222+4z2224y22=222+4y2224x22=222+4x222

4z22= 2z2x2y +4z(2z2x2y)4z22= 2z2+2- +4z(2z2+2-)

NORMALIZERS

Normalizer of H in G

Normal subgroup

H G, if g-1H g = H, for ∀g∈G

Normalizer of H in G, H<G

NG(H) ={g∈G, if g-1H g = H}

G ≥ NG(H) ≥ H

What is the normalizer NG(H) if H G?

Number of subgroups Hi<G in a conjugate class

n=[G:NG(H)]

Problem 2.13

Consider the group 4mm and its subgroups of index 4. Determine their normalizers in 4mm. Distribute the subgroups into conjugacy classes with the help of their normalizers in 4mm.


Problem 2.10 SOLUTION

{1,my}

Normalizer of {1,my} in 4mm

4mm

{1,mx}

{e,2,4,4-1,mx,my,m+,m-}

2mm={e,2,mx,my}

Conjugate subgroups: 4mm=2mm+4(2mm)

{ }

{1,my}

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Crystallographic Point Groups -